By: stewstryker

stewstryker — Thu, 03 Jan 2008 21:15:24 +0000

This minor (obvious) expansion of Laurent's version gives only the earliest consecutive month.

SELECT customer_no, min(billing_month) first_cons_month
  FROM (SELECT customer_no,
               billing_month,
               last_value(l ignore NULLS) OVER(PARTITION BY customer_no ORDER BY billing_month DESC) l,
               m
          FROM (SELECT customer_no,
                       billing_month,
                       DECODE(lag(billing_month)
                              OVER(PARTITION BY customer_no ORDER BY billing_month DESC),
                              ADD_MONTHS(billing_month, 1),
                              NULL,
                              billing_month) l,
                       MAX(billing_month) OVER(PARTITION BY customer_no) m
                  FROM cust
                 ORDER BY customer_no, billing_month DESC))
 WHERE l = m
GROUP BY customer_no;

By: Laurent Schneider

Laurent Schneider — Wed, 21 Mar 2007 12:27:04 +0000

select 
  customer_no, billing_month
from (
  select customer_no, billing_month, last_value(l ignore nulls) over (partition by customer_no order by billing_month desc ) l, m
    from (
      select customer_no, 
        billing_month, 
        decode(lag(billing_month) over (partition by customer_no order by billing_month desc),
          add_months(billing_month,1),null,
          billing_month) l,
        max(billing_month) over (partition by customer_no) m
      from cust order by customer_no, billing_month desc
    )
  )
where l=m
;

may be much faster

Comments on: How to find the last consecutive date sequence in a SQL statement

By: stewstryker

By: Laurent Schneider